Let p be the plane in space that intersects the x-axis at −5, the y-axis at −2, and the z-axis at 5. find a vector v¯¯¯ that is perpendicular to p.

To find a vector perpendicular to a given plane, is necessary to find two vectors on the plane and get the vector product of these two vectors.

Solution is:

PQ * PR = - 10×i + 10×k + 25×j

We must go through the following procedure:

Identifying the three points on the plane.

The three intersects are three points on the plane, these are:

P ( -5 , 0 , 0 ).        Q ( 0 , - 2 , 0 ).     R ( 0 , 0 , 5 )

From these points we get the vectors PQ and PR

PQ = ( 0 , -2 , 0 ) - ( -5 , 0 , 0 )

PQ = ( 0 - ( -5) , -2 - 0  , 0 - 0 )

PQ = ( 5 , - 2 , 0 )

PR = ( 0 , 0, 5 ) - ( -5 , 0, 0 )

PR = ( 5 , 0, 5 )

Now we get the vector product ( it is a perpendicular vector to PQ and PR ), and  is perpendicular to the plane.

                     i        j        k

PQ * PR =    5     -2       0      = i × (-2)×(5) + 0×k + 0×j

                    5      0       5         - [( -2)×5×k + (0)×(0)×i + (5)×(5)×j

PQ * PR = - 10×i + 10×k + 25×j

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Let P be the plane that intersects

Write the equation of the plane P:

Then

The coefficients at variables x, y and z are the coordinates of perpendicular vector to the plane. Thus

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